Expert: Paul Klarreich - 9/7/2011

Question
Questioner:Gatti  Country: South Africa  Studying:Calculus A  module covers Differentiation of different types of functions,Integration & First Order differential equations,partial derivatives and growth decay

My Question
Question in problem:
Find an equation for the tangent line to the curve y=e^2x that is perpendicular to the line x+4y=3

What I have so far:
1) I know that from equation x+4y = 3 gradient=-1/4 therefore since the tangent is perpendicular to x+4y=3 its gradient=4
2)(e^2x)dy/dx = 2e^2x

It is here where I am stuck. What should I be looking at next to calculate equation? Should I take x & Y value from x+4y=3 (which gives x = 3 and y = 3/4)into account.

Can you please advise, what I need to do next. Thank you

Answer
Country:Western Cape, South Africa
Category:Calculus
Private:No
Subject:Logarithmic differentiation
Question:

Question in problem:
Find an equation for the tangent line to the curve y=e^2x that is perpendicular to the line x+4y=3

What I have so far:
1) I know that from equation x+4y = 3 gradient=-1/4 therefore since the tangent is perpendicular to x+4y=3 its gradient=4

>>>> It's just called the slope.  Gradient is used in three dimensions.

2)(e^2x)dy/dx = 2e^2x

It is here where I am stuck. What should I be looking at next to calculate equation? Should I take x & Y value from x+4y=3 (which gives x = 3 and y = 3/4)into account.

Can you please advise, what I need to do next. Thank you
................................
Translate this sentence:

Find an equation for the tangent line to the curve y=e^2x that is perpendicular to the line x+4y=3.

to:

Find an equation for the tangent line to the curve y=e^2x that is perpendicular to the line whose slope is -1/4.

then to:

Find an equation for the tangent line to the curve y=e^2x whose slope is +4.

and to:

Find the point on the curve y=e^2x where the slope of the tangent line is +4, and then the equation of that line.

and to:

Find the point on the curve y=e^2x where dy/dx = 4, and then the equation of that line.

So if dy/dx = 2 e^(2x), solve:

2 e^(2x) = 4

e^(2x) = 2

2x = ln 2

x = 1/2 ln 2

Now get y = e^(2x):

y = e^(2x) = 2

So the slope, m = 4,   x0 = 1/2 ln 2,  and  y0 = 2.

Now use the point-slope form:

y - y0 = m(x - x0), with m, x0, y0 as above.